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Exercise :

Over the space $C[a,b]$ we define the norm $$\|f\|_2 = \sqrt{\int_a^bf(x)^2\mathrm{d}x}$$ (i) Show that if the sequence $(f_n)$ converges to $f$ with respect to $\| \cdot \|_\infty$, then it also converges to $f$ with respect to $\| \cdot \|_2$.

(ii) Is $(C[a,b], \| \cdot \|_2)$ a Banach space ?

Attempt :

(i) Let $(f_n)$ be a sequence defined over $C[a,b]$ that converges to $f$ with respect to the norm $\| \cdot \|_\infty$. Then, this means that $\exists n_0 \in \mathbb N$ :

$$\|f_n-f\|_\infty< e \Leftrightarrow \max_{x \in [a,b]}|f_n(x) - f(x)| < \epsilon \; \forall n \geq n_0$$

Now, let's consider the quantity $\| f_n - f \|_2$. This is defined as :

$$\|f_n - f\|_2 = \sqrt{\int_a^b (f_n-f)^2(x)\mathrm{d}x}=\sqrt{\int_a^b[f_n(x)-f(x)]^2\mathrm{d}x}$$

Now, the last expression can be rewritten as :

$$\sqrt{\int_a^b[f_n(x)-f(x)]^2\mathrm{d}x}=\sqrt{\int_a^b |f_n(x) - f(x)|^2\mathrm{d}x}$$

But, it is :

$$|f_n(x)-f(x)|<\max_{x \in [a,b]}|f_n(x)-f(x)| < \epsilon \; \forall n \geq n_0$$

Now, since $|f_n(x)-f(x)| \geq 0$ and $\epsilon >0$, we yield :

$$|f_n(x)-f(x)|<\epsilon \Rightarrow |f_n(x)-f(x)|^2 < \epsilon^2 \equiv \epsilon' \; \forall n\geq n_0$$

And by integrating from $a$ to $b$ since $f_n, f$ are defined over $C[a,b]$ :

$$\int_a^b|f_n(x)-f(x)|^2\mathrm{d}x < \int_a^b\epsilon'\mathrm{d}x=\epsilon'(b-a) \equiv \epsilon'' \forall n\geq n_0$$

Thus, we have $\|f_n-f\|_2 < \epsilon'' \; \forall n\geq n_0$, which means that $(f_n)$ converges to $f$ with respect to the $\|\cdot\|_2$ norm.

Question : Is my approach to (i) correct ? How would one approach (ii) though ?

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Part (i) looks fine, for the second part, define the sequence, $$ f_n(x)= \begin{cases} 0,& a\leq x\leq a+\frac{b-a}{2}\\ \frac{2n}{(b-a)}(x-a-\frac{b-a}{2}),& a+\frac{b-a}{2}<x\leq a+ \frac{b-a}{2}+\frac{b-a}{2n}\\ 1,& a+\frac{b-a}{2}+\frac{b-a}{2n}<x\leq b \end{cases} $$ for which it is definitely better to draw a picture (we are taking continuous ramps between $0$ and $1$, with steeper slope, giving the ramp less and less time to get to $1$ as $n$ gets large). The point is that $f_n(x)$ is approximating the discontinuous function $\chi_{[\frac{b-a}{2},1]}(x)$ in $||\cdot||_2$, since $$ ||f_n(x)-\chi_{[\frac{b-a}{2},1]}(x)||\leq \frac{b-a}{4n} $$ where the bound comes from the area of the triangle between the two graphs (everything is $\leq 1$, so squaring only helps us). Since each $f_n$ is continuous and has a discontinuous limit in this metric, $(C([a,b]),||\cdot||_2)$ is not a Banach space.

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    $\begingroup$ Cool stuff and indeed shows it's not a Banach space BUT how does one come up with all that ? $\endgroup$ – Rebellos Oct 28 '18 at 19:57
  • $\begingroup$ I have tried to add some intuition in the post, but the point is to try and get close in this norm to the simplest discontinuous function with a continuous one. I recommend drawing a picture and then using precalculus to figure out what the $f_n$ in the picture are $\endgroup$ – operatorerror Oct 28 '18 at 19:59
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The approach to part (i) is correct. As for part (ii), Consider the sequence of functions $f_{n}(x)=I_{[a,a+\frac{1}{n}]}(x)$ .It converges in $L^2$, but is this sequence uniformly convergent?

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    $\begingroup$ This isn't an answer to the question. The question is not whether uniform convergence implies onvergence in $L^2$ $\endgroup$ – operatorerror Oct 28 '18 at 19:58
  • $\begingroup$ My mistake, I got carried away thinking about proving the non equivalence of norms. $\endgroup$ – Kesav Krishnan Oct 28 '18 at 20:05
  • $\begingroup$ It happens! Just wanted to point it out $\endgroup$ – operatorerror Oct 28 '18 at 21:30

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